Example 1.24 Consider two continuous random variables \(X\) and \(Y\) with the joint PDF given by: \[
f_{X,Y}(x,y) =
\begin{cases}
2 & \text{if } 0 < x < 1 \text{ and } 0 < y < x \\
0 & \text{otherwise}
\end{cases}
\]
We can verify that this is a valid joint PDF by checking that it is non-negative and integrates to 1 over the entire plane: \[
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy \, dx = \int_0^1 \int_0^x 2 \, dy \, dx = \int_0^1 2x \, dx = 1
\] The support of the joint distribution is the triangular region in the \(xy\)-plane where \(0 < x < 1\) and \(0 < y < x\).
The marginal distributions can be computed as follows: \[
f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy = \int_0^x 2 \, dy = 2x \quad \text{for } 0 < x < 1
\] \[
f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx = \int_y^1 2 \, dx = 2(1 - y) \quad \text{for } 0 < y < 1
\]
We can also obtain the joint cumulative distribution function (CDF) as follows. For any \(0<y<x<1\) \[
\begin{aligned}
F_{X,Y}(x,y) &= P(X \leq x, Y \leq y)\\
& = \int_{-\infty}^x \int_{-\infty}^y f_{X,Y}(u,v) \, dv \, du\\
&=\int_{0}^y \int_{0}^u f_{X,Y}(u,v) \, dv \, du+\int_{y}^x \int_{0}^y f_{X,Y}(u,v) \, dv \, du\\
&=\int_{0}^y \int_{0}^u 2 \, dv \, du+\int_{y}^x \int_{0}^y 2 \, dv \, du\\
&= \int_{0}^y 2u \, du+\int_{y}^x 2y \, du\\
&= y^2 + 2y(x-y) \\
\end{aligned}
\]
The marginals CDFs can be computed as follows:
\[
F_X(x) = \lim_{y \to \infty} F_{X,Y}(x,y) = F_{X,Y}(x,y) = x^2 + 2x(x-x) = x^2 \quad \text{for } 0 < x < 1
\]
\[
F_Y(y) = \lim_{x \to \infty} F_{X,Y}(x,y) = F_{X,Y}(1,y) = y^2 + 2y(1-y) = 2y - y^2 \quad \text{for } 0 < y < 1
\]
or, alternatively, we can compute the marginal CDFs directly from the marginal PDFs as follows:
\[
F_X(x) = \int_{-\infty}^x f_X(u) \, du = \int_0^x 2u \, du = x^2 \quad \text{for } 0 < x < 1
\] \[
F_Y(y) = \int_{-\infty}^y f_Y(v) \, dv = \int_0^y 2(1 - v) \, dv = 2y - y^2 \quad \text{for } 0 < y < 1
\]
Now consider another joint PDF given by:
\[
f_{X,Y}(x,y) =
\begin{cases}
6xy^2 & \text{if } 0 < x < 1 \text{ and } 0 < y < 1 \\
0 & \text{otherwise}
\end{cases}
\] We can verify that this is a valid joint PDF by checking that it is non-negative and integrates to 1 over the entire plane: \[
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy \, dx = \int_0^1 \int_0^1 6xy^2 \, dy \, dx = \int_0^1 2x \, dx = 1
\] The support of the joint distribution is the unit square in the \(xy\)-plane where \(0 < x < 1\) and \(0 < y < 1\).
The marginal distributions can be computed as follows:
\[
f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy = \int_0^1 6xy^2 \, dy = 2x \quad \text{for } 0 < x < 1
\] \[
f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx = \int_0^1 6xy^2 \, dx = 3y^2 \quad \text{for } 0 < y < 1
\] Then we have
\[
f_{X,Y}(x,y)=f_X(x)f_Y(y)
\]
so the random variables \(X\) and \(Y\) are independent. THis also implies that the joint cumulative distribution function (CDF) is given by: \[
F_{X,Y}(x,y) = F_X(x) F_Y(y)
\] where \(F_X(x)\) and \(F_Y(y)\) are the marginal CDFs of \(X\) and \(Y\), respectively. We can compute these marginal CDFs as follows: \[
F_X(x) = \int_{-\infty}^x f_X(u) \, du = \int_0^x 2u \, du = x^2 \quad \text{for } 0 < x < 1
\] \[
F_Y(y) = \int_{-\infty}^y f_Y(v) \, dv = \int_0^y 3v^2 \, dv = y^3 \quad \text{for } 0 < y < 1
\] Therefore, the joint CDF is given by: \[
F_{X,Y}(x,y) = F_X(x) F_Y(y) = x^2 y^3 \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1
\]